Maxwell Equations MCQs ( Electromagnetic Theory ) MCQs – Up To Date Electromagnetic Theory MCQs
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Electromagnetic Theory MCQs – Maxwell Equations MCQs ( Electromagnetic Theory ) MCQs
The most occurred mcqs of Maxwell Equations MCQs ( Electromagnetic Theory ) in past papers. Past papers of Maxwell Equations MCQs ( Electromagnetic Theory ) Mcqs. Past papers of Maxwell Equations MCQs ( Electromagnetic Theory ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Maxwell Equations MCQs ( Electromagnetic Theory ) Mcqs. The Important series of Maxwell Equations MCQs ( Electromagnetic Theory ) Mcqs are given below:
Maxwell Law 1
1. The first Maxwell law is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Faraday and Lenz law
Answer: d
Explanation: The first Maxwell equation states that Curl(E) = -dB/dt. It is based on the emf concept. Thus it is derived from the Faraday and Lenz law.
2. The benefit of Maxwell equation is that
a) Any parameter can be calculated
b) Antenna can be designed
c) Polarisation of the wave can be calculated
d) Transmission line constants can be found
Answer: a
Explanation: The Maxwell equation relates the parameters E, D, H, B. When one parameter is known the other parameters can be easily calculated. In other words, it is used to relate an electric field parameter with its equivalent magnetic field.
3. The correct sequence to find H, when D is given is
a) D-E-B-H
b) D-B-E-H
c) It cannot be computed from the data given
d) D-H
Answer: a
Explanation: There is no direct relation between D and H, so the option D-H is not possible. Using the formula D = εE, the parameter E can be computed from D. By Maxwell equation, Curl(E) = -dB/dt, the parameter B can be calculated. Using the formula B = μH, the parameter H can be calculated. Thus the sequence is D-E-B-H.
4. The curl of the electric field intensity is
a) Conservative
b) Rotational
c) Divergent
d) Static
Answer: b
Explanation: The curl of electric field intensity is Curl(E). From Maxwell law, the curl of E is a non-zero value. Thus E will be rotational.
5. Which of the following identities is always zero for static fields?
a) Grad(Curl V)
b) Curl(Div V)
c) Div(Grad V)
d) Curl(Grad V)
Answer: d
Explanation: The curl of gradient of a vector is always zero. This is because the gradient of V is E and the curl of E is zero for static fields.
6. Find the Maxwell first law value for the electric field intensity is given by A sin wt az
a) 0
b) 1
c) -1
d) A
Answer: a
Explanation: The value of Maxwell first equation is Curl(E). The curl of E is zero. Thus for the given field, the value of Maxwell equation is zero. Thus the field is irrotational.
7. Find the electric field applied on a system with electrons having a velocity 5m/s subjected to a magnetic flux of 3.6 units.
a) 15
b) 18
c) 1.38
d) 0.72
Answer: b
Explanation: The electric field intensity is the product of the velocity and the magnetic flux density. Thus E = v x B, on substituting v = 5 and B = 3.6, we get E = 5 x 3.6 = 18 units.
8. Which of the following relations holds good?
a) Bq = ILE
b) E = ILBq
c) Eq = ILB
d) B = ILEq
Answer: c
Explanation: The force of a electrostatic field in given by F = Eq. The force on a conductor is given by F = BIL. In the case when a charge exists on a conductor, both the forces can be equated. Thus Eq = BIL is true.
9. When the Maxwell equation is expressed in frequency domain, then which substitution is possible?
a) d/dt = w/j
b) d/dt = j/w
c) d/dt = jw
d) Expression in frequency domain is not possible
Answer: c
Explanation: The conversion of time to frequency domain in Maxwell equation is given by the Fourier Transform. Differentiation in time gives jw in frequency domain. Thus d/dt = jw in frequency domain.
10. Calculate the emf of a material having a flux linkage of 2t2 at time t = 1second.
a) 2
b) 4
c) 8
d) 16
Answer: b
Explanation: The emf of a material is given by Vemf = -dλ/dt. On substituting λ = 2t2, the emf is 4t. At t = 1 sec, the emf will be 4 units.
11. Calculate the emf of a material having flux density 5sin t in an area of 0.5 units.
a) 2.5 sin t
b) -2.5 cos t
c) -5 sin t
d) 5 cos t
Answer: d
Explanation: The emf can be written as Vemf = -d(∫B.ds)/dt. It can be written as Vemf = -B= -5sin t, since the integration and differentiation gets cancelled.
12. To find D from B, sequence followed will be
a) B-E-D
b) B-H-D
c) E-H-D
d) E-B-D
Answer: a
Explanation: Using Maxwell equation, from B we can calculate E by Curl(E) = -dB /dt. From E, D can be calculated by D = εE. Thus the sequence is B->E->D.
Maxwell Law 2
1. Maxwell second equation is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Coulomb law
Answer: a
Explanation: The second Maxwell equation is based on Ampere law. It states that the field intensity of a system is same as the current enclosed by it, i.e, Curl(H) = J.
2. The Maxwell second equation that is valid in any conductor is
a) Curl(H) = Jc
b) Curl(E) = Jc
c) Curl(E) = Jd
d) Curl(H) = Jd
Answer: a
Explanation: For conductors, the conductivity parameter σ is significant and only the conduction current density exists. Thus the component J = Jc and Curl(H) = Jc.
3. In dielectric medium, the Maxwell second equation becomes
a) Curl(H) = Jd
b) Curl(H) = Jc
c) Curl(E) = Jd
d) Curl(E) = Jd
Answer: a
Explanation: In dielectric medium conductivity σ will be zero. So the current density has only the displacement current density. Thus the Maxwell equation will be Curl(H) = Jd.
4. Find the displacement current density of a material with flux density of 5sin t
a) 2.5cos t
b) 2.5sin t
c) 5cos t
d) 5sin t
Answer: c
Explanation: The displacement current density is the derivative of the flux density. Thus Jd = dD/dt. Put D = 5sin t in the equation, we get Jd = 5cos t units.
5. Find the conduction current density of a material with conductivity 200units and electric field 1.5 units.
a) 150
b) 30
c) 400/3
d) 300
Answer: d
Explanation: The conduction current density is given by Jc = σE, where σ = 200 and E = 1.5. Thus we get, Jc = 200 x 1.5 = 300 units.
6. Calculate the conduction density of a material with resistivity of 0.02 units and electric intensity of 12 units.
a) 300
b) 600
c) 50
d) 120
Answer: b
Explanation: The conduction density is given by Jc = σE, where σ is the inverse of resistivity and it is 1/0.02 = 50. Thus we get, Jc = 50 x 12 = 600 units.
Magnetic Forces And Materials MCQs
7. In the conversion of line integral of H into surface integral, which theorem is used?
a) Green theorem
b) Gauss theorem
c) Stokes theorem
d) It cannot be converted
Answer: c
Explanation: To convert line integral to surface integral, i.e, in this case from line integral of H to surface integral of J, we use the Stokes theorem. Thus the Maxwell second equation can be written as ∫H.dl = ∫∫J.ds.
8. An implication of the continuity equation of conductors is given by
a) J = σ E
b) J = E/σ
c) J = σ/E
d) J = jwEσ
Answer: a
Explanation: The continuity equation indicates the current density in conductors. This is the product of the conductivity of the conductor and the electric field subjected to it. Thus J = σE is the implication of the continuity equation for conductors.
9. Find the equation of displacement current density in frequency domain.
a) Jd = jwεE
b) Jd = jwεH
c) Jd = wεE/j
d) Jd = jεE/w
Answer: a
Explanation: The displacement current density is Jd = dD/dt. Since D = εE and in frequency domain d/dt = jw, thus we get Jd = jwεE.
10. The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field intensity.
a) 0.5i – 0.5j + 0.5k
b) 0.5i + j -1.5k
c) i – j + k
d) i + j – k
Answer: b
Explanation: By Maxwell second equation, the curl of H is same as the sum of conduction current density and displacement current density. Thus Curl(H) = J = 0.5i + j – 1.5k units.
11. At dc field, the displacement current density will be
a) 0
b) 1
c) Jc
d) ∞
Answer: a
Explanation: The DC field refers to zero frequency. The conduction current is independent of the frequency, whereas the displacement current density is dependent on the frequency, i.e, Jd = jwεE. Thus at DC field, the displacement current density will be zero.
12. Both the conduction and displacement current densities coexist in which medium?
a) Only conductors in air
b) Only dielectrics in air
c) Conductors placed in any dielectric medium
d) Both the densities can never coexist
Answer: c
Explanation: Conduction density exists only for good conductors and displacement density is for dielectrics in any medium at high frequency. Thus both coexist when a conductor is placed in a dielectric medium.
Maxwell Law 3
1. The charge density of a electrostatic field is given by
a) Curl of E
b) Divergence of E
c) Curl of D
d) Divergence of D
Answer: d
Explanation: From the Gauss law for electric field, the volume charge density is the divergence of the electric flux density of the field. Thus Div(D) = ρv.
2. In the medium of free space, the divergence of the electric flux density will be
a) 1
b) 0
c) -1
d) Infinity
Answer: b
Explanation: In free space or air, the charge density will be zero. In other words, the conduction is possible in mere air medium. By gauss law, since the charge density is same as the divergence of D, the Div(D) in air/free space will be zero.
3. In a medium other than air, the electric flux density will be
a) Solenoidal
b) Curl free
c) Irrotational
d) Divergent
Answer: d
Explanation: In any medium other than the air, the conduction is possible, due to the charge carriers. Thus charge density is also non-zero. We can write from Gauss law that Div(D) is non-zero. When the divergence is said to be non-zero, the field is not solenoidal or called as divergent field.
4. For a solenoidal field, the surface integral of D will be,
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: For a solenoidal field, the divergence will be zero. By divergence theorem, the surface integral of D and the volume integral of Div(D) is same. So as the Div(D) is zero for a solenoidal field, the surface integral of D is also zero.
5. In a dipole, the Gauss theorem value will be
a) 1
b) 0
c) -1
d) 2
Answer: b
Explanation: The Gauss theorem for an electric field is given by Div(D)= ρ. In a dipole only static charge exists and the divergence will be zero. Thus the Gauss theorem value for the dipole will be zero.
6. Find the electric flux density of a material whose charge density is given by 12 units in a volume region of 0.5 units.
a) 12
b) 24
c) 6
d) 48
Answer: c
Explanation: By Gauss law, Div(D) = ρv. To get D, integrate the charge density given. Thus D = ∫ρv dv, where ρv = 12 and ∫dv = 0.5. We get, D = 12 x 0.5 = 6 units.
7. From the Gauss law for electric field, we can compute which of the following parameters?
a) B
b) H
c) E
d) A
Answer: c
Explanation: From the Gauss law for electric field, we can find the electric flux density directly. On substituting, D= ε E, the electric field intensity can be calculated.
8. The charge density of a system with the position vector as electric flux density is
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: The divergence of the electric flux density is the charge density. For a position vector xi + yj + zk, the divergence will be 1 + 1 + 1 = 3. Thus by Gauss law, the charge density is also 3.
9. The sequence for finding E when charge density is given is
a) E-D-ρv
b) E-B-ρv
c) E-H-ρv
d) E-V-ρv
Answer: a
Explanation: From the given charge density ρv, we can compute the electric flux density by Gauss law. Since, D = εE, the electric field intensity can also be computed. Thus the sequence is E-D-ρv.
10. The Gauss law employs which theorem for the calculation of charge density?
a) Green theorem
b) Stokes theorem
c) Gauss theorem
d) Maxwell equation
Answer: c
Explanation: The Gauss divergence theorem is given by ∫ D.ds = ∫Div(D).dv. From the theorem value, we can compute the charge density. Thus Gauss law employs the Gauss divergence theorem.
Maxwell Law 4
1. Which quantity is solenoidal in the electromagnetic theory?
a) Electric field intensity
b) Electric flux density
c) Magnetic field intensity
d) Magnetic flux density
Answer: d
Explanation: The divergence of the magnetic flux density is zero. This is the Maxwell fourth equation. As the divergence is zero, the quantity will be solenoidal or divergent less.
2. Which equation will be true, if the medium is considered to be air?
a) Curl(H) = 0
b) Div(H) = 0
c) Grad(H) = 0
d) Div(H) = 1
Answer: b
Explanation: From the Gauss law for magnetic field, the divergence of the magnetic flux density is zero. Also B = μH. Thus divergence of H is also zero, i.e, Div(H) = 0 is true.
3. Find the sequence to find B when E is given.
a) E-D-H-B
b) B-E-D
c) H-B-E-D
d) V-E-B
Answer: a
Explanation: From E, D can be computed as D = εE. Using the Ampere law, H can be computed from D. Finally, B can be calculated from H by B = μH.
4. The Gauss law for magnetic field is valid in
a) Air
b) Conductor
c) Dielectric
d) All cases
Answer: d
Explanation: The Gauss law for magnetic field states that the divergence of B is always zero. This is valid for all cases like free space, dielectric medium etc.
5. The sequence for finding H from E is
a) E-B-H
b) E-V-H
c) E-D-H
d) E-A-H
Answer: a
Explanation: From E, we can compute B using the Maxwell first law. Using B, the parameter H can be found since B = μH. Thus the sequence is E-B-H is true.
6. The reason for non existence of magnetic monopoles is
a) The magnetic field cannot be split
b) Due to permeability
c) Due to magnetization
d) Due to magnetostriction
Answer: a
Explanation: Practically monopoles do not exist, due to the connection between north and south poles. But theoretically, they exist. The reason for their non- existence practically is that, the magnetic field confined to two poles cannot be split or confined to a single pole.
7. The non existence of the magnetic monopole is due to which operation?
a) Gradient
b) Divergence
c) Curl
d) Laplacian
Answer: b
Explanation: The Maxwell fourth law or the Gauss law for magnetic field states that the divergence of B is zero, implies the non existence of magnetic monopoles. Thus the operation involved is divergence.
8. Will dielectric breakdown lead to formation of magnetic monopole?
a) Yes
b) No
Answer: b
Explanation: When dielectric breakdown occurs, the material loses its dielectric property and becomes a conductor. When it is subjected to a magnetic field, north and south flux lines coexists, giving magnetic force. Thus there exists magnetic dipole. Suppose if the conductor is broken into very small pieces, still there exist a magnetic dipole in every broken part. In other words, when a piece is broken into half, there cannot exist a north pole in one half and a south pole in the other. Thus monopoles never exist.
9. Which equation will hold good for a magnetic material?
a) Line integral of H is zero
b) Surface integral of H is zero
c) Line integral of B is zero
d) Surface integral of B is zero
Answer: d
Explanation: We know that the divergence of B is zero. From Stokes theorem, the surface integral of B is equal to the volume integral of divergence of B. Thus surface integral of B is also zero.
10. The dipole formation in a magnet is due to
a) Interaction between the north and south poles together
b) Interaction between the north pole with the air
c) Interaction between the south pole with the air
d) Interaction of north and south pole separately with air
Answer: a
Explanation: In any magnetic material or magnet, the dipoles exist. This is due to the magnetic lines of force joining the north to south poles. The interaction between these two poles together leads to dipole formation.
Maxwell Law in Time Static Fields
1. Calculate the emf in a material with flux linkage of 3.5t2 at 2 seconds.
a) 3.5
b) -7
c) -14
d) 28
Answer: c
Explanation: The emf induced in a material with flux linkage is given by Vemf = -dλ/dt. On substituting λ= 3.5t2, we get emf = -7t. At time t = 2sec, the emf will be -14 units.
2. Find the emf induced in a coil of 60 turns with a flux rate of 3 units.
a) -60
b) -180
c) 60
d) 180
Answer: b
Explanation: The emf induced is the product of the turns and the flux rate. Thus Vemf = -Ndφ/dt. On substituting N = 60 and dφ/dt = 3, we get emf as -60 x 3 = -180 units.
3. Find the electric field intensity of a charge 2.5C with a force of 3N.
a) -7.5
b) 7.5
c) 2.5/3
d) 3/2.5
Answer: d
Explanation: The electric field intensity is the electric force per unit charge. It is given by E = F/q. On substituting F = 2.5 and q = 3, we get E = 3/2.5 units.
4. The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is
a) 0.28
b) 28
c) 280
d) 10/2.8
Answer: b
Explanation: The electric field is the product of the velocity and the magnetic flux density given by E = v x B. On substituting v = 10 and B = 2.8, we get E = 10 x 2.8 = 28 units.
5. The line integral of the electric field intensity is
a) Mmf
b) Emf
c) Electric potential
d) Magnetic potential
Answer: b
Explanation: From the Maxwell first law, the transformer emf is given by the line integral of the electric field intensity. Thus the emf is given by ∫ E.dl.
6. Which of the following relations is correct?
a) MMF = ∫ B.dl
b) MMF = ∫ H.dl
c) EMF = ∫ E.dl
d) EMF = ∫ D.dl
Answer: c
Explanation: The emf induced in a material is given by the line integral of the electric field intensity. Thus EMF = ∫ E.dl is the correct relation.
7. For static fields, the curl of E will be
a) Rotational
b) Irrotational
c) Solenoidal
d) Divergent
Answer: b
Explanation: For static fields, the charges will be constant and the field is constant. Thus curl of the electric field intensity will be zero. This implies the field is irrotational.
8. The line integral of which parameter is zero for static fields?
a) E
b) H
c) D
d) B
Answer: a
Explanation: The field is irrotational for static fields. Thus curl of E is zero. From Stokes theorem, the line integral of E is same as the surface integral of the curl of E. Since it is zero, the line integral of E will also be zero.
9. The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be
a) 1.414
b) 1.732
c) -1.414
d) -1.732
Answer: b
Explanation: From the Ampere circuital law, the curl of H is the conduction current density. The curl of H = yi + zj + xk is –i – j – k. Thus conduction current density is –i – j – k. The magnitude will be √(1 + 1 + 1) = √3 = 1.732 units.
10. The charge density of a field with a position vector as electric flux density is given by
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div(D) = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.
Maxwell Law in Time Varying Fields
1. Find the curl of E when B is given as 15t.
a) 15
b) -15
c) 7.5
d) -7.5
Answer: b
Explanation: From Maxwell first law, we get Curl of E as the negative derivative of B with respect to time. Thus Curl(E) = -dB/dt. On substituting B= 15t and differentiating, Curl(E) = -15 units.
2. The charge build up in a capacitor is due to
a) Conduction current density
b) Displacement current density
c) Polarisation
d) Magnetization
Answer: b
Explanation: The capacitor consists of a dielectric placed between two conducting plates, subjected to a field. The current due to a dielectric is always due to the displacement current density.
3. The surface integral of which parameter is zero?
a) E
b) D
c) B
d) H
Answer: c
Explanation: The divergence of the magnetic flux density is always zero. By Stokes theorem, the surface integral of B is same as the volume integral of the divergence of B. Thus the surface integral of B is also zero.
4. Harmonic electromagnetic fields refer to fields varying sinusoidally with respect to time. State True/False.
a) True
b) False
Answer: a
Explanation: Fields that varying sinusoidally with respect to time are called as harmonic fields. An example for harmonic fields is A sin wt.
5. When electric potential is null, then the electric field intensity will be
a) 0
b) 1
c) dA/dt
d) –dA/dt
Answer: d
Explanation: The electric field intensity is given by E = -Grad(V)- dA/dt, where V is the electric potential and A is the magnetic vector potential. When V is zero, then E = -dA/dt.
6. The gradient of the magnetic vector potential can be expressed as
a) –με dV/dt
b) +με dE/dt
c) –με dA/dt
d) +με dB/dt
Answer: a
Explanation: The gradient of A is the ratio of the negative gradient of electric potential to the speed of light c. We can write c = 1/√(με). Thus grad(A) = -με dV/dt is the required expression.
7. Find the time constant of a capacitor with capacitance of 2 microfarad having an internal resistance of 4 megaohm.
a) 2
b) 0.5
c) 8
d) 0.25
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.
8. Which components exist in an electromagnetic wave?
a) Only E
b) Only H
c) Both E and H
d) Neither E or H
Answer: c
Explanation: In an electromagnetic wave, the electric and magnetic components coexist. They propagate perpendicular to each other and to the direction of propagation in space.
9. The propagation of the electromagnetic waves can be illustrated by
a) Faraday law
b) Ampere law
c) Flemming rule
d) Coulomb law
Answer: c
Explanation: By Flemming’s rule, when the thumb and the middle finger represent the inputs (say current and field respectively), then the fore finger represents the output (force, in this case). The EM propagation can be illustrated by this rule.
10. Which one of the following laws will not contribute to the Maxwell’s equations?
a) Gauss law
b) Faraday law
c) Ampere law
d) Curie Weiss law
Answer: d
Explanation: The Gauss law, Faraday law and the Ampere law are directly used to find the parameters E, H, D, B. Thus it contributes to the Maxwell equations. The Curie Weiss law pertains to the property of any magnetic material. Thus it is not related to the Maxwell equation.
Loss Tangent
1. The loss tangent refers to the
a) Power due to propagation in conductor to that in dielectric
b) Power loss
c) Current loss
d) Charge loss
Answer: a
Explanation: The loss tangent is the tangent angle formed by the plot of conduction current density vs displacement current density. It is the ratio of Jc by Jd. It represents the loss of power due to propagation in a dielectric, when compared to that in a conductor.
2. Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.
a) 22.5
b) 4.5/5
c) 5/4.5
d) 9.5
Answer: c
Explanation: The conduction current density is the product of the conductivity and the electric field. The resistivity is the reciprocal of the conductivity. Thus the required formula is Jc = σ E = E/ρ = 5/4.5 units.
3. At high frequencies, which parameter is significant?
a) Conduction current
b) Displacement current
c) Attenuation constant
d) Phase constant
Answer: b
Explanation: The conduction current occurs in metals and is independent of the frequency. The attenuation and phase constant highly depend on the varying frequency. The displacement current occurs due to dielectrics and is significant only at very high frequencies.
4. Find the loss tangent of a material with conduction current density of 5 units and displacement current density of 10 units.
a) 2
b) 0.5
c) 5
d) 10
Answer: b
Explanation: The loss tangent is the ratio of Jc by Jd. On substituting for Jc = 5 and Jd = 10, the loss tangent, tan δ = 5/10 = 0.5. It is to be noted that it is tangent angle, so that the maxima and minima lies between 1 and -1 respectively.
5. The loss tangent is also referred to as
a) Attenuation
b) Propagation
c) Dissipation factor
d) Polarization
Answer: c
Explanation: The loss tangent is the measure of the loss of power due to propagation in a dielectric, when compared to that in a conductor. Hence it is also referred to as dissipation factor.
6. The loss tangent of a wave propagation with an intrinsic angle of 20 degree is
a) Tan 20
b) Tan 40
c) Tan 60
d) Tan 80
Answer: b
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn = tan 2(20) = tan 40.
7. The expression for the loss tangent is given by
a) σ/ωε
b) ωε/σ
c) σ/ω
d) ω/ε
Answer: a
Explanation: The conduction current density is Jc = σ E and the displacement current density is Jd = jωεE. Its magnitude will be ωεE. Thus the loss tangent tan δ = Jc /Jd = σ/ωε is the required expression.
8. Find the loss angle in degrees when the loss tangent is 1.
a) 0
b) 30
c) 45
d) 90
Answer: c
Explanation: The loss tangent is tan δ, where δ is the loss angle. Given that loss tangent tan δ = 1. Thus we get δ = tan-1(1) = 450.
9. The complex permittivity is given by 2-j. Find the loss tangent.
a) 1/2
b) -1/2
c) 2
d) -2
Answer: a
Explanation: The loss tangent for a given complex permittivity of ε = ε’ – jε’’ is given by tan δ = ε’’/ ε’. Thus the loss tangent is 1/2.
10. The intrinsic angle of the wave with a loss angle of 60 is
a) 120
b) 60
c) 90
d) 30
Answer: d
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn. We get θn = δ/2 = 60/2 = 30 degrees.